Question 25

∆PQR is right angled at Q. If sinP = 12/13, then what is the value of tanR ?

Solution

Given : $$\sin P$$ = $$\frac{12}{13}$$

Also, $$\sin P=\frac{QR}{PR}=\frac{12}{13}$$

Let QR = 12 cm and PR = 13 cm

Thus, in $$\triangle$$ PQR, => $$(PQ)^2=(PR)^2-(QR)^2$$

=> $$(PQ)^2=(13)^2-(12)^2$$

=> $$(PQ)^2=169-144=25$$

=> $$PQ=\sqrt{25}=5$$ cm

To find : $$\tan R=\frac{PQ}{QR}$$

= $$\frac{5}{12}$$

=> Ans - (D)

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